Prove that $y^2\frac{\partial z}{\partial x}+xy\frac{\partial z}{\partial y}=xz $ Let $f: \mathbb R \to \mathbb R$ differentiable, $z: \mathbb R^2 \to \mathbb R$ with $z(x,y)=yf(x^2-y^2)$.
Prove that $$y^2\frac{\partial z}{\partial x}+xy\frac{\partial z}{\partial y}=xz $$
I know that $$y^2\frac{\partial z}{\partial x}=y^2y\frac{\partial}{\partial x}f(x^2-y^2)=y^3\frac{\partial}{\partial x}f(x^2-y^2)$$
and
$$xy\frac{\partial z}{\partial y}=xy[f(x^2-y^2)+y\frac{\partial}{\partial y}f(x^2-y^2)]$$
but now I am stuck. Can anyone help me with that?
 A: HINT
with short notation
$$z_x=yf'2x \implies y^2z_x=2xy^3f'$$
$$z_y=f+yf'(-2y) \implies xyz_y=xyf-2xy^3f'$$
thus

$$y^2z_x + xyz_y=xyf=xz \quad \square$$

A: The key is: Chain Rule.
Think of the function $f$ (that appears in the expression for $z$) as $f(t)$, where $t=x^2-y^2$. Then:
$$\frac{\partial}{\partial x}\left[f(x^2-y^2)\right]=\frac{df}{dt}\cdot\frac{\partial}{\partial x}(x^2-y^2)=2x\cdot\frac{df}{dt}=2x\cdot f'(x^2-y^2).$$
Similarly you can find the partial derivative in your other calculation.
A: $$z=yf(X) \quad\text{with}\quad X=x^2-y^2 \quad \begin{cases} \frac{\partial X}{\partial x}=2x \\ \frac{\partial X}{\partial y}=-2y \end{cases}$$
$\frac{\partial z}{\partial x}=yf'(X)\frac{\partial X}{\partial x}=(yf'(X))(2x)=2xyf'(X)$
$\frac{\partial z}{\partial y}=f(X)+yf'(X)\frac{\partial X}{\partial y}=f(X)+(yf'(X))(-2y)=f(X)-2y^2f'(X)$
$y^2\frac{\partial z}{\partial x}+xy\frac{\partial z}{\partial y}=y^2\left( 2xyf'(X)\right)+xy\left(f(X)-2y^2f'(X) \right)$
After simplification : $\quad y^2\frac{\partial z}{\partial x}+xy\frac{\partial z}{\partial y}=xyf(X)$
$$\quad y^2\frac{\partial z}{\partial x}+xy\frac{\partial z}{\partial y}=xz$$
